"Routine and crude calculations show that, if $n$ is sufficiently large and $3 \leq \ell \leq r-1$, then $F_\ell < F_3 + F_{r-1}$"

It looks like $F_\ell$ is first decreasing and then increasing, such that it is either bounded by $F_3$ or $F_{r-1}$. However, I do no not see how to show this formally? Especially in the context, where $r = r(n) = \mathcal{O}(n^{\frac13})$ and $p = p(n), 0 < p < 1$ such that

Assume n is large and fixed, and also fix r, possibly needing 3r < n. Then see how the quantity varies with l. Gerhard "Ask Me About System Design" Paseman, 2011.03.11
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Gerhard PasemanMar 11 '11 at 17:57

1 Answer
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If I remember correctly, I convinced myself of a very similar statement in the Alon/Spencer book (but there even the sum of the intermediate $F_l$ had to be dominated by $F_3+F_{r-1}$) by looking at the quotient

For small $l$ (something like $l\leqslant\log_{1/p}n$) the quotient is less than 1 (so $F_l$ is decreasing in $l$), and for large $l$ the quotient is increasing. So in any case it's sufficient to look at the extreme values of $l$.